prove that the identity map $i:S^n\to S^n$ is not nullhomotopic. Using the fact that: For all $n\in \mathbb{N}$ there is no retraction $r:B^{n+1} \to S^n$, prove that the identity map $i:S^n\to S^n$ is not nullhomotopic.
This is a problem in section 56 of Munkres' book which I do not know how to approach, I'd be thankful for any helpful comments/answers. 
 A: A map $S^n\to X$ is null-homotopic if and only if it can be extended to the entire ball $B^{n+1}$ (you should have proved this earlier in your course). Let $\tilde{i}\colon B^{n+1}\to S^n$ be the extension of $i$ to the ball and let $f$ be the inclusion of $S^n\hookrightarrow B^{n+1}$ and consider the composition $\tilde{i}\circ f$. What can you say about this composition, and what does that say about $\tilde{i}$?
A: If $h:S^n\times [0,1]\to S^n$ is a homotopy between $i$ and a constant $c$. let $p:S^n\times [0,1]\to B^{n+1}$ defined by $p(x,t)=(1-t)x$ ir is easy to show that $p$ is continuous surjective and closed and $p(S^n\times\{1\})=0$, hence it induce an isomorphism ( the induced map is injective) $\tilde{p}:S^n\times [0,1]/(S^n\times\{1\})\to B^{n+1}$ ($\tilde{p}(x,t)=p(x,t)$), since $h$ is  constant in $S^n\times\{1\}$ it induce by $p$ a continuous map $r:B^{n+1}\to S^n$ ($r\circ p=h$), now for all $x\in S^n$; $r(x)=r(p(x))=h(x)=x$ this show that $r$ is a retraction(contradiction).
